2.1.57 problem 57

Internal problem ID [8195]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 57
Date solved : Sunday, November 10, 2024 at 03:08:32 AM
CAS classification : [[_2nd_order, _quadrature]]

Solve

\begin{align*} y^{\prime \prime }&=k \end{align*}

Solved as second order ode quadrature

Time used: 0.026 (sec)

The ODE can be written as

\[ y^{\prime \prime } = k \]

Integrating once gives

\[ y^{\prime }= k t + c_1 \]

Integrating again gives

\[ y= \frac {k \,t^{2}}{2} + c_1 x + c_2 \]

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= \frac {1}{2} k \,t^{2}+c_1 t +c_2 \\ \end{align*}

Figure 2.131: Slope field plot
\(y^{\prime \prime } = k\)

Solved as second order linear constant coeff ode

Time used: 0.093 (sec)

This is second order non-homogeneous ODE. In standard form the ODE is

\[ A y''(t) + B y'(t) + C y(t) = f(t) \]

Where \(A=1, B=0, C=0, f(t)=k\). Let the solution be

\[ y = y_h + y_p \]

Where \(y_h\) is the solution to the homogeneous ODE \( A y''(t) + B y'(t) + C y(t) = 0\), and \(y_p\) is a particular solution to the non-homogeneous ODE \(A y''(t) + B y'(t) + C y(t) = f(t)\). \(y_h\) is the solution to

\[ y^{\prime \prime } = 0 \]

This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

\[ A y''(t) + B y'(t) + C y(t) = 0 \]

Where in the above \(A=1, B=0, C=0\). Let the solution be \(y=e^{\lambda t}\). Substituting this into the ODE gives

\[ \lambda ^{2} {\mathrm e}^{t \lambda } = 0 \tag {1} \]

Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda t}\) gives

\[ \lambda ^{2} = 0 \tag {2} \]

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula

\[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \]

Substituting \(A=1, B=0, C=0\) into the above gives

\begin{align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {\left (0\right )^2 - (4) \left (1\right )\left (0\right )}\\ &= 0 \end{align*}

Hence this is the case of a double root \(\lambda _{1,2} = 0\). Therefore the solution is

\[ y= c_1 1 + c_2 t \tag {1} \]

Therefore the homogeneous solution \(y_h\) is

\[ y_h = c_2 t +c_1 \]

The particular solution is now found using the method of undetermined coefficients. Looking at the RHS of the ode, which is

\[ 1 \]

Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is

\[ [\{1\}] \]

While the set of the basis functions for the homogeneous solution found earlier is

\[ \{1, t\} \]

Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The UC_set becomes

\[ [\{t\}] \]

Since \(t\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The UC_set becomes

\[ [\{t^{2}\}] \]

Since there was duplication between the basis functions in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis function in the above updated UC_set.

\[ y_p = A_{1} t^{2} \]

The unknowns \(\{A_{1}\}\) are found by substituting the above trial solution \(y_p\) into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives

\[ 2 A_{1} = k \]

Solving for the unknowns by comparing coefficients results in

\[ \left [A_{1} = \frac {k}{2}\right ] \]

Substituting the above back in the above trial solution \(y_p\), gives the particular solution

\[ y_p = \frac {k \,t^{2}}{2} \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left (c_2 t +c_1\right ) + \left (\frac {k \,t^{2}}{2}\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= \frac {1}{2} k \,t^{2}+c_2 t +c_1 \\ \end{align*}

Figure 2.132: Slope field plot
\(y^{\prime \prime } = k\)

Solved as second order linear exact ode

Time used: 0.058 (sec)

An ode of the form

\begin{align*} p \left (t \right ) y^{\prime \prime }+q \left (t \right ) y^{\prime }+r \left (t \right ) y&=s \left (t \right ) \end{align*}

is exact if

\begin{align*} p''(t) - q'(t) + r(t) &= 0 \tag {1} \end{align*}

For the given ode we have

\begin{align*} p(x) &= 1\\ q(x) &= 0\\ r(x) &= 0\\ s(x) &= k \end{align*}

Hence

\begin{align*} p''(x) &= 0\\ q'(x) &= 0 \end{align*}

Therefore (1) becomes

\begin{align*} 0- \left (0\right ) + \left (0\right )&=0 \end{align*}

Hence the ode is exact. Since we now know the ode is exact, it can be written as

\begin{align*} \left (p \left (t \right ) y^{\prime }+\left (q \left (t \right )-p^{\prime }\left (t \right )\right ) y\right )' &= s(x) \end{align*}

Integrating gives

\begin{align*} p \left (t \right ) y^{\prime }+\left (q \left (t \right )-p^{\prime }\left (t \right )\right ) y&=\int {s \left (t \right )\, dt} \end{align*}

Substituting the above values for \(p,q,r,s\) gives

\begin{align*} y^{\prime }&=\int {k\, dt} \end{align*}

We now have a first order ode to solve which is

\begin{align*} y^{\prime } = k t +c_1 \end{align*}

Since the ode has the form \(y^{\prime }=f(t)\), then we only need to integrate \(f(t)\).

\begin{align*} \int {dy} &= \int {k t +c_1\, dt}\\ y &= \frac {1}{2} k \,t^{2}+c_1 t + c_2 \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= \frac {1}{2} k \,t^{2}+c_1 t +c_2 \\ \end{align*}

Figure 2.133: Slope field plot
\(y^{\prime \prime } = k\)

Solved as second order missing y ode

Time used: 0.052 (sec)

This is second order ode with missing dependent variable \(y\). Let

\begin{align*} p(t) &= y^{\prime } \end{align*}

Then

\begin{align*} p'(t) &= y^{\prime \prime } \end{align*}

Hence the ode becomes

\begin{align*} p^{\prime }\left (t \right )-k = 0 \end{align*}

Which is now solve for \(p(t)\) as first order ode. Since the ode has the form \(p^{\prime }\left (t \right )=f(t)\), then we only need to integrate \(f(t)\).

\begin{align*} \int {dp} &= \int {k\, dt}\\ p \left (t \right ) &= k t + c_1 \end{align*}

For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = k t +c_1 \end{align*}

Since the ode has the form \(y^{\prime }=f(t)\), then we only need to integrate \(f(t)\).

\begin{align*} \int {dy} &= \int {k t +c_1\, dt}\\ y &= \frac {1}{2} k \,t^{2}+c_1 t + c_2 \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= \frac {1}{2} k \,t^{2}+c_1 t +c_2 \\ \end{align*}

Figure 2.134: Slope field plot
\(y^{\prime \prime } = k\)

Solved as second order integrable as is ode

Time used: 0.036 (sec)

Integrating both sides of the ODE w.r.t \(t\) gives

\begin{align*} \int y^{\prime \prime }d t &= \int k d t\\ y^{\prime } = k t + c_1 \end{align*}

Which is now solved for \(y\). Since the ode has the form \(y^{\prime }=f(t)\), then we only need to integrate \(f(t)\).

\begin{align*} \int {dy} &= \int {k t +c_1\, dt}\\ y &= \frac {1}{2} k \,t^{2}+c_1 t + c_2 \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= \frac {1}{2} k \,t^{2}+c_1 t +c_2 \\ \end{align*}

Figure 2.135: Slope field plot
\(y^{\prime \prime } = k\)

Solved as second order integrable as is ode (ABC method)

Time used: 0.035 (sec)

Writing the ode as

\[ y^{\prime \prime } = k \]

Integrating both sides of the ODE w.r.t \(t\) gives

\begin{align*} \int y^{\prime \prime }d t &= \int k d t\\ y^{\prime } = k t +c_1 \end{align*}

Which is now solved for \(y\). Since the ode has the form \(y^{\prime }=f(t)\), then we only need to integrate \(f(t)\).

\begin{align*} \int {dy} &= \int {k t +c_1\, dt}\\ y &= \frac {1}{2} k \,t^{2}+c_1 t + c_2 \end{align*}

Will add steps showing solving for IC soon.

Figure 2.136: Slope field plot
\(y^{\prime \prime } = k\)

Solved as second order can be made integrable

Time used: 0.471 (sec)

Multiplying the ode by \(y^{\prime }\) gives

\[ y^{\prime } y^{\prime \prime }-y^{\prime } k = 0 \]

Integrating the above w.r.t \(t\) gives

\begin{align*} \int \left (y^{\prime } y^{\prime \prime }-y^{\prime } k \right )d t &= 0 \\ \frac {{y^{\prime }}^{2}}{2}-k y &= c_1 \end{align*}

Which is now solved for \(y\). Solving for the derivative gives these ODE’s to solve

\begin{align*} \tag{1} y^{\prime }&=\sqrt {2 k y+2 c_1} \\ \tag{2} y^{\prime }&=-\sqrt {2 k y+2 c_1} \\ \end{align*}

Now each of the above is solved separately.

Solving Eq. (1)

Integrating gives

\begin{align*} \int \frac {1}{\sqrt {2 k y +2 c_1}}d y &= dt\\ \frac {\sqrt {2 k y +2 c_1}}{k}&= t +c_2 \end{align*}

Singular solutions are found by solving

\begin{align*} \sqrt {2 k y +2 c_1}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = -\frac {c_1}{k} \end{align*}

Solving for \(y\) gives

\begin{align*} y &= -\frac {-c_2^{2} k^{2}-2 c_2 \,k^{2} t -k^{2} t^{2}+2 c_1}{2 k} \\ \end{align*}

Solving Eq. (2)

Integrating gives

\begin{align*} \int -\frac {1}{\sqrt {2 k y +2 c_1}}d y &= dt\\ -\frac {\sqrt {2 k y +2 c_1}}{k}&= t +c_3 \end{align*}

Singular solutions are found by solving

\begin{align*} -\sqrt {2 k y +2 c_1}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = -\frac {c_1}{k} \end{align*}

Solving for \(y\) gives

\begin{align*} y &= -\frac {-c_3^{2} k^{2}-2 c_3 \,k^{2} t -k^{2} t^{2}+2 c_1}{2 k} \\ \end{align*}

Will add steps showing solving for IC soon.

The solution

\[ y = -\frac {c_1}{k} \]

was found not to satisfy the ode or the IC. Hence it is removed.

Summary of solutions found

\begin{align*} y &= -\frac {-c_2^{2} k^{2}-2 c_2 \,k^{2} t -k^{2} t^{2}+2 c_1}{2 k} \\ y &= -\frac {-c_3^{2} k^{2}-2 c_3 \,k^{2} t -k^{2} t^{2}+2 c_1}{2 k} \\ \end{align*}

Figure 2.137: Slope field plot
\(y^{\prime \prime } = k\)

Solved as second order ode using Kovacic algorithm

Time used: 0.121 (sec)

Writing the ode as

\begin{align*} y^{\prime \prime } &= 0 \tag {1} \\ A y^{\prime \prime } + B y^{\prime } + C y &= 0 \tag {2} \end{align*}

Comparing (1) and (2) shows that

\begin{align*} A &= 1 \\ B &= 0\tag {3} \\ C &= 0 \end{align*}

Applying the Liouville transformation on the dependent variable gives

\begin{align*} z(t) &= y e^{\int \frac {B}{2 A} \,dt} \end{align*}

Then (2) becomes

\begin{align*} z''(t) = r z(t)\tag {4} \end{align*}

Where \(r\) is given by

\begin{align*} r &= \frac {s}{t}\tag {5} \\ &= \frac {2 A B' - 2 B A' + B^2 - 4 A C}{4 A^2} \end{align*}

Substituting the values of \(A,B,C\) from (3) in the above and simplifying gives

\begin{align*} r &= \frac {0}{1}\tag {6} \end{align*}

Comparing the above to (5) shows that

\begin{align*} s &= 0\\ t &= 1 \end{align*}

Therefore eq. (4) becomes

\begin{align*} z''(t) &= 0 \tag {7} \end{align*}

Equation (7) is now solved. After finding \(z(t)\) then \(y\) is found using the inverse transformation

\begin{align*} y &= z \left (t \right ) e^{-\int \frac {B}{2 A} \,dt} \end{align*}

The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of \(r\) and the order of \(r\) at \(\infty \). The following table summarizes these cases.

Case	Allowed pole order for \(r\)	Allowed value for \(\mathcal {O}(\infty )\)
1	\(\left \{ 0,1,2,4,6,8,\cdots \right \} \)	\(\left \{ \cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \right \} \)
2	Need to have at least one pole that is either order \(2\) or odd order greater than \(2\). Any other pole order is allowed as long as the above condition is satisfied. Hence the following set of pole orders are all allowed. \(\{1,2\}\),\(\{1,3\}\),\(\{2\}\),\(\{3\}\),\(\{3,4\}\),\(\{1,2,5\}\).	no condition
3	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)

Table 2.14: Necessary conditions for each Kovacic case

The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\). Therefore

\begin{align*} O\left (\infty \right ) &= \text {deg}(t) - \text {deg}(s) \\ &= 0 - -\infty \\ &= \infty \end{align*}

There are no poles in \(r\). Therefore the set of poles \(\Gamma \) is empty. Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(infinity\) then the necessary conditions for case one are met. Therefore

\begin{align*} L &= [1] \end{align*}

Since \(r = 0\) is not a function of \(t\), then there is no need run Kovacic algorithm to obtain a solution for transformed ode \(z''=r z\) as one solution is

\[ z_1(t) = 1 \]

Using the above, the solution for the original ode can now be found. The first solution to the original ode in \(y\) is found from

\[ y_1 = z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dt} \]

Since \(B=0\) then the above reduces to

\begin{align*} y_1 &= z_1 \\ &= 1 \\ \end{align*}

Which simplifies to

\[ y_1 = 1 \]

The second solution \(y_2\) to the original ode is found using reduction of order

\[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dt}}{y_1^2} \,dt \]

Since \(B=0\) then the above becomes

\begin{align*} y_2 &= y_1 \int \frac {1}{y_1^2} \,dt \\ &= 1\int \frac {1}{1} \,dt \\ &= 1\left (t\right ) \\ \end{align*}

Therefore the solution is

\begin{align*} y &= c_1 y_1 + c_2 y_2 \\ &= c_1 \left (1\right ) + c_2 \left (1\left (t\right )\right ) \\ \end{align*}

This is second order nonhomogeneous ODE. Let the solution be

\[ y = y_h + y_p \]

Where \(y_h\) is the solution to the homogeneous ODE \( A y''(t) + B y'(t) + C y(t) = 0\), and \(y_p\) is a particular solution to the nonhomogeneous ODE \(A y''(t) + B y'(t) + C y(t) = f(t)\). \(y_h\) is the solution to

\[ y^{\prime \prime } = 0 \]

The homogeneous solution is found using the Kovacic algorithm which results in

\[ y_h = c_2 t +c_1 \]

The particular solution is now found using the method of undetermined coefficients. Looking at the RHS of the ode, which is

\[ 1 \]

Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is

\[ [\{1\}] \]

While the set of the basis functions for the homogeneous solution found earlier is

\[ \{1, t\} \]

Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The UC_set becomes

\[ [\{t\}] \]

Since \(t\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The UC_set becomes

\[ [\{t^{2}\}] \]

Since there was duplication between the basis functions in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis function in the above updated UC_set.

\[ y_p = A_{1} t^{2} \]

The unknowns \(\{A_{1}\}\) are found by substituting the above trial solution \(y_p\) into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives

\[ 2 A_{1} = k \]

Solving for the unknowns by comparing coefficients results in

\[ \left [A_{1} = \frac {k}{2}\right ] \]

Substituting the above back in the above trial solution \(y_p\), gives the particular solution

\[ y_p = \frac {k \,t^{2}}{2} \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left (c_2 t +c_1\right ) + \left (\frac {k \,t^{2}}{2}\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= \frac {1}{2} k \,t^{2}+c_2 t +c_1 \\ \end{align*}

Figure 2.138: Slope field plot
\(y^{\prime \prime } = k\)

Solved as second order ode adjoint method

Time used: 0.357 (sec)

In normal form the ode

\begin{align*} y^{\prime \prime } = k \tag {1} \end{align*}

Becomes

\begin{align*} y^{\prime \prime }+p \left (t \right ) y^{\prime }+q \left (t \right ) y&=r \left (t \right ) \tag {2} \end{align*}

Where

\begin{align*} p \left (t \right )&=0\\ q \left (t \right )&=0\\ r \left (t \right )&=k \end{align*}

The Lagrange adjoint ode is given by

\begin{align*} \xi ^{''}-(\xi \, p)'+\xi q &= 0\\ \xi ^{''}-\left (0\right )' + \left (0\right ) &= 0\\ \xi ^{\prime \prime }\left (t \right )&= 0 \end{align*}

Which is solved for \(\xi (t)\). Integrating twice gives the solution

\[ \xi = c_1 t + c_2 \]

Will add steps showing solving for IC soon.

The original ode now reduces to first order ode

\begin{align*} \xi \left (t \right ) y^{\prime }-y \xi ^{\prime }\left (t \right )+\xi \left (t \right ) p \left (t \right ) y&=\int \xi \left (t \right ) r \left (t \right )d t\\ y^{\prime }+y \left (p \left (t \right )-\frac {\xi ^{\prime }\left (t \right )}{\xi \left (t \right )}\right )&=\frac {\int \xi \left (t \right ) r \left (t \right )d t}{\xi \left (t \right )} \end{align*}

Or

\begin{align*} y^{\prime }-\frac {y c_1}{c_1 t +c_2}&=\frac {k \left (\frac {1}{2} c_1 \,t^{2}+c_2 t \right )}{c_1 t +c_2} \end{align*}

Which is now a first order ode. This is now solved for \(y\). In canonical form a linear first order is

\begin{align*} y^{\prime } + q(t)y &= p(t) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(t) &=-\frac {c_1}{c_1 t +c_2}\\ p(t) &=\frac {c_1 k \,t^{2}+2 c_2 k t}{2 c_1 t +2 c_2} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dt}}\\ &= {\mathrm e}^{\int -\frac {c_1}{c_1 t +c_2}d t}\\ &= \frac {1}{c_1 t +c_2} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}}\left ( \mu y\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {c_1 k \,t^{2}+2 c_2 k t}{2 c_1 t +2 c_2}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \left (\frac {y}{c_1 t +c_2}\right ) &= \left (\frac {1}{c_1 t +c_2}\right ) \left (\frac {c_1 k \,t^{2}+2 c_2 k t}{2 c_1 t +2 c_2}\right ) \\ \mathrm {d} \left (\frac {y}{c_1 t +c_2}\right ) &= \left (\frac {c_1 k \,t^{2}+2 c_2 k t}{2 \left (c_1 t +c_2 \right )^{2}}\right )\, \mathrm {d} t \\ \end{align*}

Integrating gives

\begin{align*} \frac {y}{c_1 t +c_2}&= \int {\frac {c_1 k \,t^{2}+2 c_2 k t}{2 \left (c_1 t +c_2 \right )^{2}} \,dt} \\ &=\frac {k \left (\frac {t}{c_1}+\frac {c_2^{2}}{c_1^{2} \left (c_1 t +c_2 \right )}\right )}{2} + c_3 \end{align*}

Dividing throughout by the integrating factor \(\frac {1}{c_1 t +c_2}\) gives the final solution

\[ y = \left (c_1 t +c_2 \right ) \left (\frac {k \left (\frac {t}{c_1}+\frac {c_2^{2}}{c_1^{2} \left (c_1 t +c_2 \right )}\right )}{2}+c_3 \right ) \]

Hence, the solution found using Lagrange adjoint equation method is

\begin{align*} y &= \left (c_1 t +c_2 \right ) \left (\frac {k \left (\frac {t}{c_1}+\frac {c_2^{2}}{c_1^{2} \left (c_1 t +c_2 \right )}\right )}{2}+c_3 \right ) \\ \end{align*}

The constants can be merged to give

\[ y = \left (c_1 t +c_2 \right ) \left (\frac {k \left (\frac {t}{c_1}+\frac {c_2^{2}}{c_1^{2} \left (c_1 t +c_2 \right )}\right )}{2}+1\right ) \]

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= \left (c_1 t +c_2 \right ) \left (\frac {k \left (\frac {t}{c_1}+\frac {c_2^{2}}{c_1^{2} \left (c_1 t +c_2 \right )}\right )}{2}+1\right ) \\ \end{align*}

Figure 2.139: Slope field plot
\(y^{\prime \prime } = k\)

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right )=k \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right ) \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {0}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=1 \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (t \right )\hspace {3pt}\textrm {by}\hspace {3pt} t \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=t \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=\mathit {C1} y_{1}\left (t \right )+\mathit {C2} y_{2}\left (t \right )+y_{p}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=\mathit {C1} +\mathit {C2} t +y_{p}\left (t \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} y_{p}\left (t \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Use variation of parameters to find}\hspace {3pt} y_{p}\hspace {3pt}\textrm {here}\hspace {3pt} f \left (t \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [y_{p}\left (t \right )=-y_{1}\left (t \right ) \left (\int \frac {y_{2}\left (t \right ) f \left (t \right )}{W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )}d t \right )+y_{2}\left (t \right ) \left (\int \frac {y_{1}\left (t \right ) f \left (t \right )}{W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )}d t \right ), f \left (t \right )=k \right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=\left [\begin {array}{cc} 1 & t \\ 0 & 1 \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=1 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=k \left (-\left (\int t d t \right )+\left (\int 1d t \right ) t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {k \,t^{2}}{2} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=\mathit {C1} +\mathit {C2} t +\frac {1}{2} k \,t^{2} \end {array} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
<- quadrature successful`
 

Maple dsolve solution

Solving time : 0.002 (sec)
Leaf size : 15

dsolve(diff(diff(y(t),t),t) = k, 
       y(t),singsol=all)
 

\[ y = \frac {1}{2} k \,t^{2}+c_{1} t +c_{2} \]

Mathematica DSolve solution

Solving time : 0.003 (sec)
Leaf size : 20

DSolve[{D[y[t],{t,2}]==k,{}}, 
       y[t],t,IncludeSingularSolutions->True]
 

\[ y(t)\to \frac {k t^2}{2}+c_2 t+c_1 \]